7b - Solution: The height of the block initially is: h =...

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PHY2053 – Section 7392 Quiz #7 Friday, February 27, 2004 Name_______ KEY ___________________ UFID___________________ Please read the problem carefully and be sure to show all of your work. Partial credit will be given. Starting from rest, an 18.0 kg block slides down a frictionless ramp inclined at 25° with respect to the floor to the bottom. The block then slides an additional 3.0 along the floor before coming to a stop. Determine the coefficient of friction between the block and the floor and the mechanical energy lost due to friction.
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Unformatted text preview: Solution: The height of the block initially is: h = 2.0 sin 25 = 0.845 m The gravitational potential energy at the top is: mgh = 149.2 J At the bottom of the ramp, all of the energy is converted to kinetic energy. However, the block loses all of its kinetic energy due to friction, eventually coming to rest after a distance d . Thus: mgh = Fd But F = N= mg Thus, mgh = mgd and solving for : = h/d= 0.28 Note that you do not need to compute the kinetic energy!...
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This note was uploaded on 09/25/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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7b - Solution: The height of the block initially is: h =...

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